Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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+ | {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #19]] and [[2013 AMC 10B Problems|2013 AMC 10B #23]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | A | + | In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=14</math>, and <math>CA=15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD}\perp\overline{BC}</math>, <math>\overline{DE}\perp\overline{AC}</math>, and <math>\overline{AF}\perp\overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? |
+ | |||
+ | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | size(200); | ||
+ | defaultpen(linewidth(0.4)+fontsize(10)); | ||
+ | pen s = linewidth(0.8)+fontsize(8); | ||
+ | |||
+ | pair A,B,C,D,E0,F,G; | ||
+ | A = origin; | ||
+ | C = (15,0); | ||
+ | B = IP(CR(A,13),CR(C,14)); | ||
+ | D = foot(A,C,B); | ||
+ | E0 = foot(D,A,C); | ||
+ | F = OP(CR((A+B)/2,length(B-A)/2), D--E0); | ||
+ | draw(A--C--B--A, black+0.8); | ||
+ | draw(B--F--A--D--E0); | ||
+ | dot("$A$",A,W); | ||
+ | dot("$B$",B,N); | ||
+ | dot("$C$",C,E); | ||
+ | dot("$D$",D,NE); | ||
+ | dot("$E$",E0,S); | ||
+ | dot("$F$",F,E); | ||
+ | draw(rightanglemark(B,D,A,15)); | ||
+ | draw(rightanglemark(B,F,A,15)); | ||
+ | draw(rightanglemark(D,E0,A,15)); | ||
+ | label("$5$",D--B,NE); | ||
+ | label("$9$",D--C,NE); | ||
+ | label(Label("$13$",Rotate(B-A)), B--A); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 1== | ||
+ | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>, so <math>\triangle ABF \sim \triangle ADE</math> are similar. In addition, <math>\triangle ADE \sim \triangle ACD</math>. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using Pythagorean triples. | ||
+ | |||
+ | So, the ratio of the longer leg to the hypotenuse of all three similar triangles is <math>\tfrac{12}{15} = \tfrac{4}{5}</math>, and the ratio of the shorter leg to the hypotenuse is <math>\tfrac{9}{15} = \tfrac{3}{5}</math>. It follows that <math>AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)</math>. | ||
+ | |||
+ | Let <math>x=DF</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <cmath>13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.</cmath> Dividing by <math>13</math> we get <math>x+4=7.2\implies x=\frac{16}{5}</math> so our answer is <math>\boxed{\textbf{(B) }21}</math>. | ||
+ | |||
+ | ~Edits by BakedPotato66 | ||
+ | |||
+ | ==Solution 2== | ||
+ | From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{27}{5} = \frac{48}{5}</math> . Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = 180 - \angle{DFA} = \angle{EFA}</math>, and triangles <math>\triangle AEF \sim \triangle ADB</math>. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math>. <math>m + n = 16 + 5 = 21</math> and our answer is <math>\boxed{\textbf{(B) } 21}</math>. | ||
+ | |||
+ | <math>\triangle ADC \sim \triangle DEC</math> because of <math>AA \sim</math> . <math>\angle{ADC} = \angle{DEC} = 90°</math>. Lets say <math>\angle{ADE} = x</math>. So <math>\angle{EDC} = 90 - x</math> and <math>\angle{DEC} = 180 - 90 - (90 - x) = x</math> so <math>\angle{ADE} = \angle{DEC}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ||
+ | |||
+ | ==Solution 3== | ||
+ | If we draw a diagram as given, but then add point <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{FG}\perp\overline{BC}</math> in order to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG=\tfrac35x</math> and <math>DG=\tfrac45x</math>, where <math>x</math> is the length of <math>\overline{DF}</math>. Using the Pythagorean theorem, we now get <cmath>BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}</cmath> and <math>AF</math> can be found out noting that <math>AE</math> is just <math>\tfrac{48}5</math> through base times height (since <math>12\cdot 9 = 15 \cdot \tfrac{36}5</math>, similar triangles gives <math>AE = \tfrac{48}5</math>), and that <math>EF</math> is just <math>\tfrac{36}5 - x</math>. From there, <cmath>AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.</cmath> Now, <math>BF^2 + AF^2 = 169</math>, and squaring and adding both sides and subtracting a 169 from both sides gives <math>2x^2 - \tfrac{32}5x = 0</math>, so <math>x = \tfrac{16}5</math>. Thus, the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | ==Solution 4 (Power of a Point)== | ||
+ | First, we find <math>BD = 5</math>, <math>DC = 9</math>, and <math>AD = 12</math> via the Pythagorean Theorem or by using similar triangles. Next, because <math>DE</math> is an altitude of triangle <math>ADC</math>, <math>DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}</math>. Using that, we can use the Pythagorean Theorem and similar triangles to find <math>EC = \frac{27}{5}</math> and <math>AE = \frac{48}{5}</math>. | ||
+ | |||
+ | Points <math>A</math>, <math>B</math>, <math>D</math>, and <math>F</math> all lie on a circle whose diameter is <math>AB</math>. Let the point where the circle intersects <math>AC</math> be <math>G</math>. Using power of a point, we can write the following equation to solve for <math>AG</math>: <cmath>DC\cdot BC = CG\cdot AC</cmath> <cmath>9\cdot 14 = CG\cdot 15</cmath> <cmath>CG = 126/15</cmath> Using that, we can find <math>AG = \frac{99}{15}</math>, and using <math>AG</math>, we can find that <math>GE = 3</math>. | ||
+ | |||
+ | We can use power of a point again to solve for <math>DF</math>: <cmath>FE\cdot DE = GE\cdot AE</cmath> <cmath>(\frac{36}{5} - DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}</cmath> <cmath>\frac{36}{5} - DF = 4</cmath> <cmath>DF = \frac{16}{5} = \frac{m}{n}</cmath> Thus, <math>m+n = 16+5 = 21</math> <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 5 (Coord Bash)== | ||
+ | If we draw the diagram like above, but make <math>BC</math> the base, we can set <math>B(0,0), C(14,0)</math> and <math>A(5,12)</math>, as <math>AD</math> can quickly be verified to be <math>12</math> by Pythagorean triples or similar triangles. Construct <math>X</math> on <math>BC</math> such that <math>EX \perp BC</math>. This implies <math>\triangle ADC \sim \triangle EXC</math> as <math>AD \parallel EX</math>, and <math> \angle ACD = \angle ECX </math>. Also construct <math>FY</math> such that <math>FY \perp BC</math>. | ||
+ | |||
+ | |||
+ | Line <math>\overline {AC}</math> has a slope of <math>-\frac{4}{3}</math> by slope formula. Since <math>D = (5,0)</math> and <math>DE \perp AC,</math>, the equation of <math> DE = \frac{3}{4}x - \frac{15}{4}</math>. | ||
+ | |||
+ | Furthermore, <math>F</math> can now be expressed as <math>(x,\frac{3}{4}x - \frac{15}{4}).</math> Since we know <math>AF \perp BF,</math> we can solve for <math>x</math> with the perpendicular slope formula like so: <cmath>\frac{\frac{3}{4}x - \frac{63}{4}}{x - 5} \cdot \frac{\frac{3}{4}x - \frac {15}{4}}{x} = -1</cmath> <cmath>\frac{x-21}{x-5} \cdot \frac{x-5}{x} = -\frac{16}{9}</cmath> <cmath>\frac{x-21}{x} = -\frac{16}{9}</cmath> <cmath> 9x = 189 -16x </cmath> <cmath>x = \frac{189}{25}</cmath> | ||
+ | |||
+ | Plugging <math>\frac{189}{25}</math> into <math>F</math>, we get <math>F(\frac{189}{25},\frac{48}{25}).</math> Since <math>FY \perp DY</math>, we get that <math>\triangle FYD</math> has side lengths of <math>FY = \frac{48}{25}</math> | ||
+ | |||
+ | and <math>DY = BY - BD = BY - 5 = \frac{64}{25}</math>. | ||
+ | |||
+ | Clearly, <math>\triangle FYD</math> is a <math>3 - 4 - 5</math> pythagorean triple, so <math>DF = \frac{80}{25} = \frac{16}{5}</math>. | ||
+ | |||
+ | <math>16 + 5 = m + n = \boxed{\textbf{(B) }21}</math>. | ||
+ | |||
+ | ~JT0543164 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/XZBKnobK-JU?t=3064 | ||
− | + | == See also == | |
+ | {{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} | ||
− | + | {{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}} | |
− | + | {{MAA Notice}} |
Latest revision as of 22:39, 29 July 2024
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
Problem
In triangle ,
,
, and
. Distinct points
,
, and
lie on segments
,
, and
, respectively, such that
,
, and
. The length of segment
can be written as
, where
and
are relatively prime positive integers. What is
?
Diagram
Solution 1
Since , quadrilateral
is cyclic. It follows that
, so
are similar. In addition,
. We can easily find
,
, and
using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is , and the ratio of the shorter leg to the hypotenuse is
. It follows that
.
Let . By Ptolemy's Theorem, we have
Dividing by
we get
so our answer is
.
~Edits by BakedPotato66
Solution 2
From solution 1, we know that and
. Since
, we can figure out that
. We also know what
is so we can figure what
is:
. Quadrilateral
is cyclic, implying that
= 180°. Therefore,
, and triangles
. Solving the resulting proportion gives
. Therefore,
.
and our answer is
.
because of
.
. Lets say
. So
and
so
Solution 3
If we draw a diagram as given, but then add point on
such that
in order to use the Pythagorean theorem, we end up with similar triangles
and
. Thus,
and
, where
is the length of
. Using the Pythagorean theorem, we now get
and
can be found out noting that
is just
through base times height (since
, similar triangles gives
), and that
is just
. From there,
Now,
, and squaring and adding both sides and subtracting a 169 from both sides gives
, so
. Thus, the answer is
.
Solution 4 (Power of a Point)
First, we find ,
, and
via the Pythagorean Theorem or by using similar triangles. Next, because
is an altitude of triangle
,
. Using that, we can use the Pythagorean Theorem and similar triangles to find
and
.
Points ,
,
, and
all lie on a circle whose diameter is
. Let the point where the circle intersects
be
. Using power of a point, we can write the following equation to solve for
:
Using that, we can find
, and using
, we can find that
.
We can use power of a point again to solve for :
Thus,
.
Solution 5 (Coord Bash)
If we draw the diagram like above, but make the base, we can set
and
, as
can quickly be verified to be
by Pythagorean triples or similar triangles. Construct
on
such that
. This implies
as
, and
. Also construct
such that
.
Line has a slope of
by slope formula. Since
and
, the equation of
.
Furthermore, can now be expressed as
Since we know
we can solve for
with the perpendicular slope formula like so:
Plugging into
, we get
Since
, we get that
has side lengths of
and .
Clearly, is a
pythagorean triple, so
.
.
~JT0543164
Video Solution
https://youtu.be/XZBKnobK-JU?t=3064
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.